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Chapter 16: Problem 16

Two wires are parallel, and one is directly above the other. Each has a length of \(50.0 \mathrm{m}\) and a mass per unit length of \(0.020 \mathrm{kg} / \mathrm{m}\). However, the tension in wire \(\mathrm{A}\) is \(6.00 \times 10^{2} \mathrm{N},\) and the tension in wire \(\mathrm{B}\) is \(3.00 \times 10^{2} \mathrm{N} .\) Transverse wave pulses are generated simultaneously, one at the left end of wire \(A\) and one at the right end of wire \(B\). The pulses travel toward each other. How much time does it take until the pulses pass each other?

Short Answer

Expert verified
It takes approximately 0.169 seconds for the pulses to pass each other.

Step by step solution

01

Determine Wave Speed in Wire A

The wave speed in a wire is given by the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the wire, and \( \mu \) is the mass per unit length. For wire A, \( T = 6.00 \times 10^2 \) N and \( \mu = 0.020 \) kg/m. Therefore, the wave speed in wire A is: \[ v_A = \sqrt{\frac{6.00 \times 10^2}{0.020}} \approx 173.205 \text{ m/s} \]
02

Determine Wave Speed in Wire B

Similarly, for wire B, use the same formula. The tension \( T = 3.00 \times 10^2 \) N and \( \mu = 0.020 \) kg/m. The wave speed in wire B is: \[ v_B = \sqrt{\frac{3.00 \times 10^2}{0.020}} \approx 122.474 \text{ m/s} \]
03

Calculate the Time to Meet

The pulses travel towards each other from opposite ends of the 50 m length. The sum of their speeds gives the relative speed of approach: \[ v_{\text{relative}} = v_A + v_B = 173.205 + 122.474 = 295.679 \text{ m/s} \] The time \( t \) it takes for the pulses to meet is the length of the wire divided by their relative speed: \[ t = \frac{50.0}{295.679} \approx 0.169 \text{ seconds} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tension in Wire
Tension is a force that stretches a wire, affecting the speed of waves traveling through it. Imagine tension like pulling a string tight; the tighter you pull, the faster vibrations can move along it. In this exercise, wire A has a tension of \(600\, \text{N}\) and wire B has \(300\, \text{N}\). This difference impacts the wave speeds in each wire.

When determining wave speed due to tension, use the formula:
  • \( v = \sqrt{\frac{T}{\mu}} \)
where \( T \) is the tension, and \( \mu \) is the mass per unit length of the wire.

Higher tension results in higher wave speed. This is why wire A's waves travel faster than those in wire B.
Mass Per Unit Length
Mass per unit length, denoted as \( \mu \), expresses how much mass is distributed over each meter of the wire. Think of it like spreading a certain weight evenly along a rope. This characteristic influences how quickly waves can move through the wire.

For both wires in the problem, \( \mu = 0.020\, \text{kg/m} \). This means each meter of wire holds 0.020 kg of mass. Having the same mass per unit length allows us to focus on how tension alone affects wave speed.

Understanding \( \mu \) is crucial because it acts together with tension in the wave speed formula. If \( \mu \) increases, wave speed decreases, since more mass means more inertia to overcome. It's like trying to wave a heavier rope.
Relative Speed of Waves
When two waves move towards each other, understanding their relative speed helps determine when and where they will meet. The relative speed is simply the sum of the speeds of the two waves.

For wire A and B, find the wave speeds using their given tensions:
  • Wave speed in wire A: \( v_A = 173.205\, \text{m/s} \)
  • Wave speed in wire B: \( v_B = 122.474\, \text{m/s} \)
The relative speed becomes \( v_{\text{relative}} = v_A + v_B = 295.679\, \text{m/s} \).

This value tells us how fast the distance between the waves closes. Since they start from opposite ends of the 50 m wires, this speed helps calculate the meeting time with:
  • Time = \( \frac{\text{Length of wire}}{v_{\text{relative}}} \approx 0.169\, \text{seconds} \)
This quick calculation shows the efficiency of waves meeting due to combined speeds.

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Most popular questions from this chapter

A wireless transmitting microphone is mounted on a small platform that can roll down an incline, directly away from a loudspeaker that is mounted at the top of the incline. The loudspeaker broadcasts a tone that has a fixed frequency of \(1.000 \times 10^{4} \mathrm{Hz},\) and the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) At a time of 1.5 s following the release of the platform, the microphone detects a frequency of \(9939 \mathrm{Hz} .\) At a time of \(3.5 \mathrm{s}\) following the release of the platform. the microphone detects a frequency of \(9857 \mathrm{Hz}\). What is the acceleration (assumed constant) of the platform?

Suppose you are part of a team that is trying to break the sound barrier with a jet-powered car, which means that it must travel faster than the speed of sound in air. In the morning, the air temperature is \(0^{\circ} \mathrm{C},\) and the speed of sound is \(331 \mathrm{m} / \mathrm{s} .\) What speed must your car exceed if it is to break the sound barrier when the temperature has risen to \(43^{\circ} \mathrm{C}\) in the afternoon? Assume that air behaves like an ideal gas.

Suppose that the linear density of the A string on a violin is \(7.8 \times 10^{-4} \mathrm{kg} / \mathrm{m} .\) A wave on the string has a frequency of \(440 \mathrm{Hz}\) and a wavelength of \(65 \mathrm{cm} .\) What is the tension in the string?

Two submarines are under water and approaching each other head-on. Sub A has a speed of \(12 \mathrm{m} / \mathrm{s}\) and sub \(\mathrm{B}\) has a speed of \(8 \mathrm{m} / \mathrm{s}\). Sub \(\mathrm{A}\) sends out a \(1550-\mathrm{Hz}\) sonar wave that travels at a speed of \(1522 \mathrm{m} / \mathrm{s}\). (a) What is the frequency detected by sub \(B ?\) (b) Part of the sonar wave is reflected from sub \(B\) and returns to sub \(A\). What frequency does sub A detect for this reflected wave?

A siren, mounted on a tower, emits a sound whose frequency is \(2140 \mathrm{Hz}\). A person is driving a car away from the tower at a speed of \(27.0 \mathrm{m} / \mathrm{s}\). As the figure illustrates, the sound reaches the person by two paths: the sound reflected from the building in front of the car, and the sound coming directly from the siren. The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). Concepts: (i) One way that the Doppler effect can arise is that the wavelength of the sound changes. For either the direct or reflected sound, does the wavelength change? (ii) Why does the driver hear a frequency for the reflected sound that is different from \(2140 \mathrm{Hz},\) and is it greater or smaller than \(2140 \mathrm{Hz} ?\) (iii) Why does the driver hear a frequency for the direct sound that is different from \(2140 \mathrm{Hz}\), and is it greater or smaller than \(2140 \mathrm{Hz}\) ? Calculations: What frequency does the person hear for the (a) reflected and (b) direct sound?
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